a)\(\frac{7\cdot15:25\cdot102}{7\cdot4\cdot15\cdot3:25\cdot4\cdot102\cdot3}=\frac{1}{4\cdot3:4\cdot3}=\frac{1}{1}=1\)
b)\(\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}=\frac{2+3+4+5+6+7+8}{10}=\frac{44}{10}=\frac{22}{5}\)